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In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).
Correct answer is '2732.05'. Can you explain this answer?
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Most Upvoted Answer
In an orthogonal machining operation, the cutting and thrust forces ar...
Given : Thickness of chip (t) = 0.5 mm
Shear angle (φ)= 150
rake angle (α)= 00
Width of cut (b) = 2 mm
Yield shear strength ( τs) = 500 MPa
Cutting force = thrust force
then
Hence, the cutting force is 2732.05 N.
Shear angle (φ)= 150
rake angle (α)= 00
Width of cut (b) = 2 mm
Yield shear strength ( τs) = 500 MPa
Cutting force = thrust force
then
Hence, the cutting force is 2732.05 N.
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In an orthogonal machining operation, the cutting and thrust forces ar...
°. Determine the magnitude of the cutting and thrust forces.
To determine the magnitude of the cutting and thrust forces, we can use the following equations:
Cutting force (Fc) = uncut chip thickness (t) * shear angle (θ) * shear strength (τ)
Thrust force (Ft) = uncut chip thickness (t) * shear strength (τ)
Given:
Uncut chip thickness (t) = 0.5 mm
Shear angle (θ) = 15°
We need the shear strength (τ) to calculate the forces. The shear strength depends on the material being machined and is typically given in the problem statement or can be looked up in tables.
For example, let's assume the shear strength (τ) is 300 MPa.
Cutting force (Fc) = 0.5 mm * 15° * 300 MPa = 225 N
Thrust force (Ft) = 0.5 mm * 300 MPa = 150 N
Therefore, the magnitude of the cutting force is 225 N and the magnitude of the thrust force is 150 N.
To determine the magnitude of the cutting and thrust forces, we can use the following equations:
Cutting force (Fc) = uncut chip thickness (t) * shear angle (θ) * shear strength (τ)
Thrust force (Ft) = uncut chip thickness (t) * shear strength (τ)
Given:
Uncut chip thickness (t) = 0.5 mm
Shear angle (θ) = 15°
We need the shear strength (τ) to calculate the forces. The shear strength depends on the material being machined and is typically given in the problem statement or can be looked up in tables.
For example, let's assume the shear strength (τ) is 300 MPa.
Cutting force (Fc) = 0.5 mm * 15° * 300 MPa = 225 N
Thrust force (Ft) = 0.5 mm * 300 MPa = 150 N
Therefore, the magnitude of the cutting force is 225 N and the magnitude of the thrust force is 150 N.
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In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. Can you explain this answer?
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In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. 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 operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. 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 operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. Can you explain this answer?.
In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. 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 operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. 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 operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. Can you explain this answer?.
Solutions for In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).Correct answer is '2732.05'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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