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In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____
Correct answer is between '825,835'. Can you explain this answer?
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In an orthogonal machining experiment carried out using a cutting tool...
Rake angle (α) = 0°
Cutting force(Fc) = 1700 N
Friction angle (β) = 26°
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In an orthogonal machining experiment carried out using a cutting tool...
° and the shear angle is 15°, calculate the shear force and normal force acting on the tool.
First, we need to determine the components of the cutting force acting parallel and perpendicular to the rake face. Since the rake angle is zero, the cutting force is entirely perpendicular to the rake face. Therefore, the normal force acting on the tool is:
Normal force = Cutting force = 1700 N
To calculate the shear force, we need to resolve the cutting force into its components parallel and perpendicular to the shear plane. The friction angle at the rake face-chip interface is 26°, which means that the shear plane is inclined at an angle of 26° to the rake face. The shear angle is 15°, which means that the shear plane is inclined at an angle of 15° to the direction of motion of the tool.
Using trigonometry, we can calculate the components of the cutting force parallel and perpendicular to the shear plane. The component of the cutting force perpendicular to the shear plane is:
Perpendicular force = Cutting force x sin(26°) = 1700 x sin(26°) = 745.8 N
The component of the cutting force parallel to the shear plane is:
Parallel force = Cutting force x cos(26°) = 1700 x cos(26°) = 1536.1 N
The shear force acting on the tool is equal to the parallel force, which is 1536.1 N. Therefore, the shear force is 1536.1 N and the normal force is 1700 N.
First, we need to determine the components of the cutting force acting parallel and perpendicular to the rake face. Since the rake angle is zero, the cutting force is entirely perpendicular to the rake face. Therefore, the normal force acting on the tool is:
Normal force = Cutting force = 1700 N
To calculate the shear force, we need to resolve the cutting force into its components parallel and perpendicular to the shear plane. The friction angle at the rake face-chip interface is 26°, which means that the shear plane is inclined at an angle of 26° to the rake face. The shear angle is 15°, which means that the shear plane is inclined at an angle of 15° to the direction of motion of the tool.
Using trigonometry, we can calculate the components of the cutting force parallel and perpendicular to the shear plane. The component of the cutting force perpendicular to the shear plane is:
Perpendicular force = Cutting force x sin(26°) = 1700 x sin(26°) = 745.8 N
The component of the cutting force parallel to the shear plane is:
Parallel force = Cutting force x cos(26°) = 1700 x cos(26°) = 1536.1 N
The shear force acting on the tool is equal to the parallel force, which is 1536.1 N. Therefore, the shear force is 1536.1 N and the normal force is 1700 N.
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In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. Can you explain this answer?
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In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. 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 an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. 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 an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. Can you explain this answer?.
In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. 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 an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. 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 an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. Can you explain this answer?.
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In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. Can you explain this answer?, a detailed solution for In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. Can you explain this answer? has been provided alongside types of In an orthogonal machining experiment carried out using a cutting tool with zero degree rake angle, the measured cutting force was 1700 N. If the friction angle at the rake face-chip interface is 26°, then the thrust force value in N is _____Correct answer is between '825,835'. Can you explain this answer? theory, EduRev gives you an
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