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In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is
[ME 2007]
- a)20.56
- b)26.56
- c)30.56
- d)36.56
Correct answer is option 'B'. Can you explain this answer?
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In orthogonal turning of a low carbon steel bar of diameter 150 mm wit...
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In orthogonal turning of a low carbon steel bar of diameter 150 mm wit...
°, determine:
a) The shear angle and the shear strain rate
b) The cutting force per unit width of cut if the shear strength of the material is 500 MPa
c) The power required to perform the cutting operation if the specific energy is 2.5 J/mm3
Assuming that the chip is formed by pure shear, we can use the formula for shear angle:
tan(ϕ) = chip thickness / feed = 0.48 / 0.24 = 2
Therefore, ϕ = 63.4°
The shear strain rate can be calculated using the formula:
γ̇ = cutting velocity / (chip thickness * depth of cut) = 90 / (0.48 * 2) = 93.75 s^-1
The cutting force per unit width of cut can be calculated using the formula:
Fc = shear strength * chip thickness * depth of cut = 500 * 0.48 * 2 = 480 N/mm
The power required to perform the cutting operation can be calculated using the formula:
P = cutting force * cutting velocity / specific energy = (480 * 150 * π) * 90 / (2.5 * 10^6) = 103.7 kW
Therefore, the shear angle is 63.4°, the shear strain rate is 93.75 s^-1, the cutting force per unit width of cut is 480 N/mm, and the power required to perform the cutting operation is 103.7 kW.
a) The shear angle and the shear strain rate
b) The cutting force per unit width of cut if the shear strength of the material is 500 MPa
c) The power required to perform the cutting operation if the specific energy is 2.5 J/mm3
Assuming that the chip is formed by pure shear, we can use the formula for shear angle:
tan(ϕ) = chip thickness / feed = 0.48 / 0.24 = 2
Therefore, ϕ = 63.4°
The shear strain rate can be calculated using the formula:
γ̇ = cutting velocity / (chip thickness * depth of cut) = 90 / (0.48 * 2) = 93.75 s^-1
The cutting force per unit width of cut can be calculated using the formula:
Fc = shear strength * chip thickness * depth of cut = 500 * 0.48 * 2 = 480 N/mm
The power required to perform the cutting operation can be calculated using the formula:
P = cutting force * cutting velocity / specific energy = (480 * 150 * π) * 90 / (2.5 * 10^6) = 103.7 kW
Therefore, the shear angle is 63.4°, the shear strain rate is 93.75 s^-1, the cutting force per unit width of cut is 480 N/mm, and the power required to perform the cutting operation is 103.7 kW.
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In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. Can you explain this answer?
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In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. 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 In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. 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 In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. Can you explain this answer?.
In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. 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 In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. 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 In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is[ME 2007]a)20.56b)26.56c)30.56d)36.56Correct answer is option 'B'. Can you explain this answer?.
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