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A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3
- a)55.5
- b)56.5
- c)55.96
- d)none
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A thin cylindrical shell made of steel, of diameter 250 mm, wall thick...
Maximum stress,
σc = 
= = 120 N/mm2
= = 120 N/mm2P= 
= 5.76 N/mm2
= 5.76 N/mm2E = 200000 N/mm2 , ν = 0.3
Axial strain,
εa =
= 
= 1.2 × 10−4
Circumferential strain,
εc =
= 
= 5.1 × 10−4
Volumetric strain, εv = 2εc + εa
= 2 × 5.1 × 10−4 + 1.2 × 10−4
= 11.4 × 10−4
Original volume,
V =
=
= 49.0874 × 106 mm3
Change in volume, δV = εv × V
= 11.4 × 10−4 × 49.0874 × 106
= 55960 mm3 = 55.96 cm3
= 55.96 cc
Question_Type: 5
Most Upvoted Answer
A thin cylindrical shell made of steel, of diameter 250 mm, wall thick...
's ratio = 0.3, determine the following:
a) The magnitude of the internal pressure
b) The change in diameter of the shell due to the internal pressure
c) The change in length of the shell due to the internal pressure
Solution:
a) The maximum stress developed in the shell can be calculated using the formula:
σ = pd/2t
where σ is the maximum stress, p is the internal pressure, d is the diameter, and t is the wall thickness.
Substituting the given values, we get:
120 MPa = p(250 mm)/(2 x 6 mm)
Solving for p, we get:
p = 7.2 MPa
Therefore, the magnitude of the internal pressure is 7.2 MPa.
b) The change in diameter of the shell due to the internal pressure can be calculated using the formula:
Δd/d = -ν(p/E)
where Δd is the change in diameter, d is the original diameter, ν is the Poisson's ratio, p is the internal pressure, and E is the modulus of elasticity.
Substituting the given values, we get:
Δd/250 mm = -0.3(7.2 MPa)/(200 GPa)
Solving for Δd, we get:
Δd = -0.00216 mm
Therefore, the change in diameter of the shell due to the internal pressure is -0.00216 mm (which means the diameter decreases by this amount).
c) The change in length of the shell due to the internal pressure can be calculated using the formula:
ΔL/L = ν(p/E)
where ΔL is the change in length, L is the original length, ν is the Poisson's ratio, p is the internal pressure, and E is the modulus of elasticity.
Substituting the given values, we get:
ΔL/1000 mm = 0.3(7.2 MPa)/(200 GPa)
Solving for ΔL, we get:
ΔL = 0.000648 mm
Therefore, the change in length of the shell due to the internal pressure is 0.000648 mm (which means the length increases by this amount).
a) The magnitude of the internal pressure
b) The change in diameter of the shell due to the internal pressure
c) The change in length of the shell due to the internal pressure
Solution:
a) The maximum stress developed in the shell can be calculated using the formula:
σ = pd/2t
where σ is the maximum stress, p is the internal pressure, d is the diameter, and t is the wall thickness.
Substituting the given values, we get:
120 MPa = p(250 mm)/(2 x 6 mm)
Solving for p, we get:
p = 7.2 MPa
Therefore, the magnitude of the internal pressure is 7.2 MPa.
b) The change in diameter of the shell due to the internal pressure can be calculated using the formula:
Δd/d = -ν(p/E)
where Δd is the change in diameter, d is the original diameter, ν is the Poisson's ratio, p is the internal pressure, and E is the modulus of elasticity.
Substituting the given values, we get:
Δd/250 mm = -0.3(7.2 MPa)/(200 GPa)
Solving for Δd, we get:
Δd = -0.00216 mm
Therefore, the change in diameter of the shell due to the internal pressure is -0.00216 mm (which means the diameter decreases by this amount).
c) The change in length of the shell due to the internal pressure can be calculated using the formula:
ΔL/L = ν(p/E)
where ΔL is the change in length, L is the original length, ν is the Poisson's ratio, p is the internal pressure, and E is the modulus of elasticity.
Substituting the given values, we get:
ΔL/1000 mm = 0.3(7.2 MPa)/(200 GPa)
Solving for ΔL, we get:
ΔL = 0.000648 mm
Therefore, the change in length of the shell due to the internal pressure is 0.000648 mm (which means the length increases by this amount).
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A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. Can you explain this answer?
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A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. 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 A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. 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 A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. Can you explain this answer?.
A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. 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 A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. 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 A thin cylindrical shell made of steel, of diameter 250 mm, wall thickness 6 mm, length 1 m is subjected to internal pressure p such that maximum stress developed in the shell is 120 MPa. If E = 200 GPa and Poisson’s ratio is 0.3. The change in volume of the shell is ___________ cm3a)55.5b)56.5c)55.96d)noneCorrect answer is option 'C'. Can you explain this answer?.
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