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The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:
26.035 = VT0.13 f0.77 d0.37
A 100 mm tool life was obtained under the following cutting conditions:
Velocity = 25 m/min
Feed = 0.3125 mm/rev
Diameter of the bar = 2.50 mm
What will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)
Correct answer is '24'. Can you explain this answer?
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Most Upvoted Answer
The equation given below has been obtained while machining AISI 2340 ...
We are given
Velocity (V) = 25 min
Feed (f) = 0.3125 min/rev
Diameter of bar (d) = 2.50 mm
For a tool life of 100 min:
When V is increased by 20%; feed and diameter of the bar remain constant.
Substituting the values, we have
26.035 = V(100)0.13 ...(i)
But 26.035 = 1.2 V(T)0.13 ...(ii)
Rewriting equations (i) and (ii), we have
(V/1.2V) = (100/T)0.13 = 1
or T0.13 = (100)0.13/1.2
T0.13 = 1.82/1.2
T0.13 = 1.51
T = 24 min
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The equation given below has been obtained while machining AISI 2340 ...
Given Data:
- Tool life equation: 26.035 = VT^0.13 f^0.77 d^0.37A
- Tool signature: 8, 22, 6, 6, 15, 6, 0.117 mm
- Initial cutting conditions: Velocity = 25 m/min, Feed = 0.3125 mm/rev, Diameter = 2.50 mm
- Initial tool life: 100 min
Calculations:
1. Calculate the initial values of V, f, and d:
- Velocity (V) = 25 m/min
- Feed (f) = 0.3125 mm/rev
- Diameter (d) = 2.50 mm
2. Substitute the initial values into the tool life equation to find constant A:
26.035 = 25^0.13 * 0.3125^0.77 * 2.5^0.37 * A
A = 1.511
3. Calculate the initial tool life using the constant A:
Tool life = 1.511 * 25^0.13 * 0.3125^0.77 * 2.5^0.37
= 100 min
4. Increase the cutting speed by 20%:
New velocity = 25 m/min + 20% of 25 m/min = 30 m/min
5. Calculate the new tool life using the updated velocity:
New tool life = 1.511 * 30^0.13 * 0.3125^0.77 * 2.5^0.37
≈ 24 min
Therefore, the tool life when the cutting speed is increased by 20% will be approximately 24 minutes.
- Tool life equation: 26.035 = VT^0.13 f^0.77 d^0.37A
- Tool signature: 8, 22, 6, 6, 15, 6, 0.117 mm
- Initial cutting conditions: Velocity = 25 m/min, Feed = 0.3125 mm/rev, Diameter = 2.50 mm
- Initial tool life: 100 min
Calculations:
1. Calculate the initial values of V, f, and d:
- Velocity (V) = 25 m/min
- Feed (f) = 0.3125 mm/rev
- Diameter (d) = 2.50 mm
2. Substitute the initial values into the tool life equation to find constant A:
26.035 = 25^0.13 * 0.3125^0.77 * 2.5^0.37 * A
A = 1.511
3. Calculate the initial tool life using the constant A:
Tool life = 1.511 * 25^0.13 * 0.3125^0.77 * 2.5^0.37
= 100 min
4. Increase the cutting speed by 20%:
New velocity = 25 m/min + 20% of 25 m/min = 30 m/min
5. Calculate the new tool life using the updated velocity:
New tool life = 1.511 * 30^0.13 * 0.3125^0.77 * 2.5^0.37
≈ 24 min
Therefore, the tool life when the cutting speed is increased by 20% will be approximately 24 minutes.
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The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer?
Question Description
The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer?.
The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer?.
Solutions for The equation given below has been obtained while machining AISI 2340 steel with H.S.S. cutting tools having a 8, 22, 6, 6, 15, 6, 0.117 mm tool signature. The work tool system is governed by the following equation:26.035 = VT0.13 f0.77 d0.37A 100 mm tool life was obtained under the following cutting conditions:Velocity = 25 m/minFeed = 0.3125 mm/revDiameter of the bar = 2.50 mmWhat will be the tool life (in min) if cutting speed is increased by 20%? (Answer up to the nearest integer)Correct answer is '24'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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