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During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)
Correct answer is '20.5'. Can you explain this answer?
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During friction welding, copper bars of 8 mm are coaxially welded end...
Force (F) = Pressure x Area x μ
Torque = F x Radius
Torque = 5026 x 4 x 10-3 = 20.11 Nm
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During friction welding, copper bars of 8 mm are coaxially welded end...
Given:
- Diameter of copper bars (D) = 8 mm
- Axial pressure (P) = 200 MPa
- Rotational speed (N) = 450 rpm
- Coefficient of friction (μ) = 0.50
- Required torque at the interface of the welding = ?
Formula:
The torque required at the interface of the welding can be calculated using the formula:
Torque (T) = 2πF
Where,
F = Frictional force between the mating faces of the rotating bars
Calculating the Frictional force:
Frictional force (F) can be calculated using the formula:
F = μPπ(D/2)^2
Where,
D = Diameter of the copper bars
Given: D = 8 mm
Converting Diameter to meters:
Diameter (D) = 8 mm = 8 × 10^(-3) m
Calculating the Frictional force:
F = μPπ(D/2)^2
F = 0.50 × 200 × 10^6 × π × (8 × 10^(-3)/2)^2
Simplifying the equation:
F = 0.50 × 200 × 10^6 × π × (4 × 10^(-3))^2
F = 0.50 × 200 × 10^6 × π × 16 × 10^(-6)
F = 1 × 10^8 × π × 16 × 10^(-6)
Calculating the Torque:
Torque (T) = 2πF
T = 2π × 1 × 10^8 × π × 16 × 10^(-6)
Simplifying the equation:
T = 2π^2 × 1 × 10^8 × 16 × 10^(-6)
T = 2 × 3.14^2 × 1 × 10^8 × 16 × 10^(-6)
T = 2 × 3.14^2 × 1 × 0.16 × 10^2
T = 2 × 3.14^2 × 16
T = 2 × 9.86 × 16
T = 313.6 N m
The torque required at the interface of the welding is 313.6 N m. However, the answer provided is 20.5 N m, which is incorrect. It seems there might be a mistake in the given answer or the calculations provided.
- Diameter of copper bars (D) = 8 mm
- Axial pressure (P) = 200 MPa
- Rotational speed (N) = 450 rpm
- Coefficient of friction (μ) = 0.50
- Required torque at the interface of the welding = ?
Formula:
The torque required at the interface of the welding can be calculated using the formula:
Torque (T) = 2πF
Where,
F = Frictional force between the mating faces of the rotating bars
Calculating the Frictional force:
Frictional force (F) can be calculated using the formula:
F = μPπ(D/2)^2
Where,
D = Diameter of the copper bars
Given: D = 8 mm
Converting Diameter to meters:
Diameter (D) = 8 mm = 8 × 10^(-3) m
Calculating the Frictional force:
F = μPπ(D/2)^2
F = 0.50 × 200 × 10^6 × π × (8 × 10^(-3)/2)^2
Simplifying the equation:
F = 0.50 × 200 × 10^6 × π × (4 × 10^(-3))^2
F = 0.50 × 200 × 10^6 × π × 16 × 10^(-6)
F = 1 × 10^8 × π × 16 × 10^(-6)
Calculating the Torque:
Torque (T) = 2πF
T = 2π × 1 × 10^8 × π × 16 × 10^(-6)
Simplifying the equation:
T = 2π^2 × 1 × 10^8 × 16 × 10^(-6)
T = 2 × 3.14^2 × 1 × 10^8 × 16 × 10^(-6)
T = 2 × 3.14^2 × 1 × 0.16 × 10^2
T = 2 × 3.14^2 × 16
T = 2 × 9.86 × 16
T = 313.6 N m
The torque required at the interface of the welding is 313.6 N m. However, the answer provided is 20.5 N m, which is incorrect. It seems there might be a mistake in the given answer or the calculations provided.
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During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer?
Question Description
During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. 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 During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. 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 During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer?.
During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. 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 During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. 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 During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer?.
Solutions for During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer?, a detailed solution for During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer? has been provided alongside types of During friction welding, copper bars of 8 mm are coaxially welded end to end at an axial pressure of 200 MPa and rotational speed of 450 rpm. The coefficient of friction between the mating faces of the rotating bars is 0.50. The torque (in N m) required at the interface of the welding is . (Rounded upto two decimal places)Correct answer is '20.5'. Can you explain this answer? theory, EduRev gives you an
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