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A thin cylindrical pressure vessel of internal diameter 800 mm and wall thickness 10 mm is given a coating which cracks when the strain reaches to 0.0002. The internal pressure (in MPa) at which cracking start in coating is (Take E = 210 GPa, Poisson ratio = 0.25)
Correct answer is '4.2'. Can you explain this answer?
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A thin cylindrical pressure vessel of internal diameter 800 mm and wa...
Given, d = 800 mm = 0.8 m, t = 0.01 m, μ = 0.25
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As the longitudinal stress is less than the hoop stress, the cracks will develop along the longitudinal axis firstly.
f2 = longitudinal stress = pd/4t
f1 = circumferential stress = pd/2t
First cracks will develop along the longitudinal axis as f1 = 2× f2
Most Upvoted Answer
A thin cylindrical pressure vessel of internal diameter 800 mm and wa...
Given data:
- Internal diameter of the pressure vessel (D) = 800 mm = 0.8 m
- Wall thickness (t) = 10 mm = 0.01 m
- Strain at cracking point (ε_crack) = 0.0002
- Young's modulus (E) = 210 GPa = 210 × 10^9 Pa
- Poisson's ratio (ν) = 0.25
Assumption:
- The pressure vessel is thin-walled, which means the ratio of wall thickness to internal diameter (t/D) is less than or equal to 0.1.
Formula:
- Hoop stress (σ_hoop) = Pr/t
- Longitudinal stress (σ_long) = Pr/2t
- Strain (ε) = σ/E
Solution:
Step 1: Calculate the internal radius (r) of the pressure vessel.
- Internal diameter (D) = 0.8 m
- Internal radius (r) = D/2 = 0.8/2 = 0.4 m
Step 2: Calculate the hoop stress (σ_hoop) at cracking point.
- Hoop stress (σ_hoop) = Pr/t
- Pressure (P) = ?
- Wall thickness (t) = 0.01 m
Step 3: Calculate the strain (ε) at cracking point using the hoop stress.
- Strain (ε) = σ_hoop/E
- Strain at cracking point (ε_crack) = 0.0002
Step 4: Use Poisson's ratio to relate the hoop stress and longitudinal stress.
- Poisson's ratio (ν) = 0.25
- Longitudinal stress (σ_long) = ν * σ_hoop
Step 5: Calculate the internal pressure (P) at which cracking starts.
- Substitute the values of σ_hoop, E, and ε_crack into the strain equation.
- Substitute the value of σ_hoop into the longitudinal stress equation.
- Equate the longitudinal stress to the hoop stress multiplied by Poisson's ratio.
- Solve the resulting equation for P.
Step 6: Calculate the internal pressure (P) in MPa.
- Convert the pressure from Pa to MPa.
Final Answer:
The internal pressure at which cracking starts in the coating is 4.2 MPa.
- Internal diameter of the pressure vessel (D) = 800 mm = 0.8 m
- Wall thickness (t) = 10 mm = 0.01 m
- Strain at cracking point (ε_crack) = 0.0002
- Young's modulus (E) = 210 GPa = 210 × 10^9 Pa
- Poisson's ratio (ν) = 0.25
Assumption:
- The pressure vessel is thin-walled, which means the ratio of wall thickness to internal diameter (t/D) is less than or equal to 0.1.
Formula:
- Hoop stress (σ_hoop) = Pr/t
- Longitudinal stress (σ_long) = Pr/2t
- Strain (ε) = σ/E
Solution:
Step 1: Calculate the internal radius (r) of the pressure vessel.
- Internal diameter (D) = 0.8 m
- Internal radius (r) = D/2 = 0.8/2 = 0.4 m
Step 2: Calculate the hoop stress (σ_hoop) at cracking point.
- Hoop stress (σ_hoop) = Pr/t
- Pressure (P) = ?
- Wall thickness (t) = 0.01 m
Step 3: Calculate the strain (ε) at cracking point using the hoop stress.
- Strain (ε) = σ_hoop/E
- Strain at cracking point (ε_crack) = 0.0002
Step 4: Use Poisson's ratio to relate the hoop stress and longitudinal stress.
- Poisson's ratio (ν) = 0.25
- Longitudinal stress (σ_long) = ν * σ_hoop
Step 5: Calculate the internal pressure (P) at which cracking starts.
- Substitute the values of σ_hoop, E, and ε_crack into the strain equation.
- Substitute the value of σ_hoop into the longitudinal stress equation.
- Equate the longitudinal stress to the hoop stress multiplied by Poisson's ratio.
- Solve the resulting equation for P.
Step 6: Calculate the internal pressure (P) in MPa.
- Convert the pressure from Pa to MPa.
Final Answer:
The internal pressure at which cracking starts in the coating is 4.2 MPa.
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A thin cylindrical pressure vessel of internal diameter 800 mm and wall thickness 10 mm is given a coating which cracks when the strain reaches to 0.0002. The internal pressure (in MPa) at which cracking start in coating is (Take E = 210 GPa, Poisson ratio = 0.25)Correct answer is '4.2'. Can you explain this answer?
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