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A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.?
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A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto ...
Problem: A steel cylinder with OD 400 mm and ID 240 mm is shrunk onto another steel cylinder with OD 240 mm and ID 140 mm. The radial interference is 0.3 mm. Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Also, find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.
Solution:
Interface Pressure (pi):
The interference fit creates a contact pressure that can be found using the following equation:
pi = (E * δ) / ((1 - n^2) * (D1/D2 + 1))
Where,
E = Young's Modulus = 200 GPa
δ = radial interference = 0.3 mm
n = Poisson's Ratio = 0.3
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation, we get:
pi = (200 * 10^9 * 0.3) / ((1 - 0.3^2) * (240/400 + 1))
pi = 345.6 MPa
Therefore, the interface pressure is 345.6 MPa.
Radial Stress:
The radial stress in the smaller cylinder can be found using the following equation:
σr1 = pi * ((D2 - D1) / D1)
Where,
pi = interface pressure = 345.6 MPa
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation, we get:
σr1 = 345.6 * ((400 - 240) / 240)
σr1 = 345.6 MPa
Therefore, the radial stress in the smaller cylinder is 345.6 MPa.
The radial stress in the larger cylinder can be found using the following equation:
σr2 = pi * ((D1 - D2) / D2)
Where,
pi = interface pressure = 345.6 MPa
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation, we get:
σr2 = 345.6 * ((240 - 400) / 400)
σr2 = -207.36 MPa
Therefore, the radial stress in the larger cylinder is -207.36 MPa (negative sign indicates compressive stress).
Tangential Stress:
The tangential stress in the smaller cylinder can be found using the following equation:
σt1 = pi * ((D2^2 - D1^2) / (D2^2 + D1^2))
Where,
pi = interface pressure = 345.6 MPa
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation
Solution:
Interface Pressure (pi):
The interference fit creates a contact pressure that can be found using the following equation:
pi = (E * δ) / ((1 - n^2) * (D1/D2 + 1))
Where,
E = Young's Modulus = 200 GPa
δ = radial interference = 0.3 mm
n = Poisson's Ratio = 0.3
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation, we get:
pi = (200 * 10^9 * 0.3) / ((1 - 0.3^2) * (240/400 + 1))
pi = 345.6 MPa
Therefore, the interface pressure is 345.6 MPa.
Radial Stress:
The radial stress in the smaller cylinder can be found using the following equation:
σr1 = pi * ((D2 - D1) / D1)
Where,
pi = interface pressure = 345.6 MPa
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation, we get:
σr1 = 345.6 * ((400 - 240) / 240)
σr1 = 345.6 MPa
Therefore, the radial stress in the smaller cylinder is 345.6 MPa.
The radial stress in the larger cylinder can be found using the following equation:
σr2 = pi * ((D1 - D2) / D2)
Where,
pi = interface pressure = 345.6 MPa
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation, we get:
σr2 = 345.6 * ((240 - 400) / 400)
σr2 = -207.36 MPa
Therefore, the radial stress in the larger cylinder is -207.36 MPa (negative sign indicates compressive stress).
Tangential Stress:
The tangential stress in the smaller cylinder can be found using the following equation:
σt1 = pi * ((D2^2 - D1^2) / (D2^2 + D1^2))
Where,
pi = interface pressure = 345.6 MPa
D1 = outer diameter of the smaller cylinder = 240 mm
D2 = outer diameter of the larger cylinder = 400 mm
Substituting the values in the above equation
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A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.?
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A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.? 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 A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.?.
A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.? 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 A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.?.
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A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.?, a detailed solution for A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.? has been provided alongside types of A 400 mm OD steel cylinder with a nominal ID of 240 mm is shrunk onto another steel cylinder of 240 mm OD and 140 mm ID. The radial interference δ is 0.3 mm. Use Young's Modulus E = 200 GPa and Poisson's Ratio n = 0.3. Find the interface pressure pi and plot the radial and tangential stresses in both cylinders. Then find the maximum internal pressure which may be applied to the assembly if the maximum tangential stress in the inside cylinder is to be no more than 140 MPa.? theory, EduRev gives you an
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