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A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.?
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A steel cylinder of 10 cm internal diameter and 16 cm external diamete...
Problem statement:
A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylinder that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm2.
Solution:
Given data:
Internal diameter of the steel cylinder, d1 = 10 cm
External diameter of the steel cylinder, d2 = 16 cm
Inside diameter of the cylinder to be shrunk, d3 = 15.25 cm
Contact pressure after shrinkage, p = 200 kg/cm2
Modulus of elasticity, E = 2*106 kg/cm2
Assumptions:
The length of the cylinder is constant before and after shrinkage.
The shrinkage of the outer cylinder is uniform along its length.
The material of both cylinders is homogeneous and isotropic.
Calculations:
The radial stress on the inner cylinder, σ1 = p(d2-d3)/(d2+d3)
The radial stress on the outer cylinder, σ2 = p(d3-d1)/(d2+d3)
The tangential stress on the inner cylinder, τ1 = p(d2+d3)/(d2-d3)
The tangential stress on the outer cylinder, τ2 = p(d2+d3)/(d2-d1)
The change in diameter of the outer cylinder after shrinkage, Δd = -σ2*d2/E
The final external diameter of the outer cylinder after shrinkage, D = d2 + Δd
Substituting the given values in the above equations, we get:
σ1 = 344.83 kg/cm2
σ2 = 126.09 kg/cm2
τ1 = 800 kg/cm2
τ2 = 1187.5 kg/cm2
Δd = -1.008 mm
D = 15.992 cm
Therefore, the external diameter of the outer cylinder after shrinkage should be 15.992 cm to achieve a contact pressure of 200 kg/cm2.
Conclusion:
The external diameter of the outer cylinder after shrinkage can be calculated by using the formula Δd = -σ2*d2/E and adding it to the original external diameter of the cylinder. Careful consideration of assumptions is necessary to ensure accurate results.
A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylinder that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm2.
Solution:
Given data:
Internal diameter of the steel cylinder, d1 = 10 cm
External diameter of the steel cylinder, d2 = 16 cm
Inside diameter of the cylinder to be shrunk, d3 = 15.25 cm
Contact pressure after shrinkage, p = 200 kg/cm2
Modulus of elasticity, E = 2*106 kg/cm2
Assumptions:
The length of the cylinder is constant before and after shrinkage.
The shrinkage of the outer cylinder is uniform along its length.
The material of both cylinders is homogeneous and isotropic.
Calculations:
The radial stress on the inner cylinder, σ1 = p(d2-d3)/(d2+d3)
The radial stress on the outer cylinder, σ2 = p(d3-d1)/(d2+d3)
The tangential stress on the inner cylinder, τ1 = p(d2+d3)/(d2-d3)
The tangential stress on the outer cylinder, τ2 = p(d2+d3)/(d2-d1)
The change in diameter of the outer cylinder after shrinkage, Δd = -σ2*d2/E
The final external diameter of the outer cylinder after shrinkage, D = d2 + Δd
Substituting the given values in the above equations, we get:
σ1 = 344.83 kg/cm2
σ2 = 126.09 kg/cm2
τ1 = 800 kg/cm2
τ2 = 1187.5 kg/cm2
Δd = -1.008 mm
D = 15.992 cm
Therefore, the external diameter of the outer cylinder after shrinkage should be 15.992 cm to achieve a contact pressure of 200 kg/cm2.
Conclusion:
The external diameter of the outer cylinder after shrinkage can be calculated by using the formula Δd = -σ2*d2/E and adding it to the original external diameter of the cylinder. Careful consideration of assumptions is necessary to ensure accurate results.
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A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.?
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A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.? 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 steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.?.
A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.? 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 steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.?.
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A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.?, a detailed solution for A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.? has been provided alongside types of A steel cylinder of 10 cm internal diameter and 16 cm external diameter is strengthened by shrinking another cylinder of the same length on to it. The inside diameter of this cylinder was originally 15.25 cm. Find the external diameter of the outer cylindo that the contact pressure after shrinkage will be 200 kg/cm2, take E= 2*10*6 kg/cm 2.? theory, EduRev gives you an
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