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Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).
Correct answer is '2573.40'. Can you explain this answer?
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Most Upvoted Answer
Under orthogonal cutting condition, a turning operation is carried out...
Given: -
Rake angle (aα)= 50
Uncut chip thickness (t1) = 0.2 mm
Friction angle (λ)= 450
Shear angle (φ)= 250
Shear stress (τ)= 1000 MPa
Width (b) = 3 mm
Shear force
FS = 1419.72
From merchant circle
Fcutting = cos(45 - 5) x 3359.34
Fcutting = 2573.40 N
Hence, the correct answer is 2573.40
Rake angle (aα)= 50
Uncut chip thickness (t1) = 0.2 mm
Friction angle (λ)= 450
Shear angle (φ)= 250
Shear stress (τ)= 1000 MPa
Width (b) = 3 mm
Shear force
FS = 1419.72
From merchant circle
Fcutting = cos(45 - 5) x 3359.34
Fcutting = 2573.40 N
Hence, the correct answer is 2573.40
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Community Answer
Under orthogonal cutting condition, a turning operation is carried out...
Given data:
Cutting speed (V) = 4 m/s
Orthogonal rake angle (α) = 50°
Uncut chip thickness (t) = 0.2 mm
Width of cut (b) = 3 mm
Friction angle (φ) = 45°
Shear angle (λ) = 25°
Dynamic yield shear strength (τy) = 1000 MPa
Calculation of shear plane angle (β):
The shear plane angle (β) can be calculated using the formula:
tan β = tan (λ - φ)
Substituting the given values:
tan β = tan (25° - 45°)
tan β = tan (-20°)
β = -20°
Calculation of shear plane area (A):
The shear plane area (A) can be calculated using the formula:
A = t * b * cos β
Substituting the given values:
A = 0.2 mm * 3 mm * cos(-20°)
A = 0.6 mm^2 * cos(-20°)
A = 0.6 mm^2 * 0.9397
A = 0.5638 mm^2
Calculation of cutting force (Fc):
The cutting force (Fc) can be calculated using the formula:
Fc = A * τy
Substituting the given values:
Fc = 0.5638 mm^2 * 1000 MPa
Fc = 563.8 N
However, the given answer is 2573.40 N, which means that the units or values used in the given answer are not consistent with the given data. Therefore, the given answer is incorrect.
To calculate the correct cutting force, the units of the given data should be consistent. If the uncut chip thickness (t) and width of cut (b) are in millimeters, then the cutting speed (V) should also be in millimeters per second. If the dynamic yield shear strength (τy) is in MPa, then the shear plane area (A) should be in square millimeters.
Assuming the units are consistent, the correct cutting force can be calculated as follows:
Fc = A * τy
Fc = 0.5638 mm^2 * 1000 MPa
Fc = 563.8 N (rounded off to one decimal place)
Therefore, the correct cutting force is 563.8 N, not 2573.40 N.
Cutting speed (V) = 4 m/s
Orthogonal rake angle (α) = 50°
Uncut chip thickness (t) = 0.2 mm
Width of cut (b) = 3 mm
Friction angle (φ) = 45°
Shear angle (λ) = 25°
Dynamic yield shear strength (τy) = 1000 MPa
Calculation of shear plane angle (β):
The shear plane angle (β) can be calculated using the formula:
tan β = tan (λ - φ)
Substituting the given values:
tan β = tan (25° - 45°)
tan β = tan (-20°)
β = -20°
Calculation of shear plane area (A):
The shear plane area (A) can be calculated using the formula:
A = t * b * cos β
Substituting the given values:
A = 0.2 mm * 3 mm * cos(-20°)
A = 0.6 mm^2 * cos(-20°)
A = 0.6 mm^2 * 0.9397
A = 0.5638 mm^2
Calculation of cutting force (Fc):
The cutting force (Fc) can be calculated using the formula:
Fc = A * τy
Substituting the given values:
Fc = 0.5638 mm^2 * 1000 MPa
Fc = 563.8 N
However, the given answer is 2573.40 N, which means that the units or values used in the given answer are not consistent with the given data. Therefore, the given answer is incorrect.
To calculate the correct cutting force, the units of the given data should be consistent. If the uncut chip thickness (t) and width of cut (b) are in millimeters, then the cutting speed (V) should also be in millimeters per second. If the dynamic yield shear strength (τy) is in MPa, then the shear plane area (A) should be in square millimeters.
Assuming the units are consistent, the correct cutting force can be calculated as follows:
Fc = A * τy
Fc = 0.5638 mm^2 * 1000 MPa
Fc = 563.8 N (rounded off to one decimal place)
Therefore, the correct cutting force is 563.8 N, not 2573.40 N.
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Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer?
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Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. 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 Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. 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 Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer?.
Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. 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 Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. 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 Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer?.
Solutions for Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer?, a detailed solution for Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer? has been provided alongside types of Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 50 . The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 450 and 250 , respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).Correct answer is '2573.40'. Can you explain this answer? tests, examples and also practice Mechanical Engineering tests.
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