5. Thin-Walled Pressure Vessels
Cylindrical and spherical vessels are commonly used in various engineering applications. These vessels are used to contain fluids.
Compared to the diameter of the vessel, the wall of the vessel itself is very thin. Typically, the wall of the vessel is \(\frac{1}{10}^{th}\) to \(\frac{1}{15}^{th}\) of its diameter. When empty, the vessel is subjected to equal atmospheric pressure inside as well as outside. However, the wall of this vessel is subjected to tensile stresses when the vessel is filled with fluid, say, compressed gas.
5.1. Failure of Thin-Walled Vessels
A thin-walled pressure vessel experiencing tensile stresses may fail in the following ways if the failure limit is exceeded.
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Split up in a longitudinal direction, along the length
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Split up in across the cross-section, along the circumference.
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Example 1 |
A cylindrical vessel \(2\ m\) long and \(500\ mm\) in diameter with plates \(10\ mm\) thick is subjected to an internal pressure of \(3\ MPa\). Taking \(E = 200\ GPa\) and Poisson’s ratio \(= 0.3\) for the vessel material, calculate the circumferential and longitudinal strain. Calculate the original and change in volume of the vessel. |
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Solution:
circumferential strain
\[\begin{aligned}
\varepsilon_{c} &=\frac{p d}{2 t E}\left(1-\frac{1}{2 m}\right)\\
%&=\frac{3 \times 500}{2 \times 10 \times\left(200 \times 10^{3}\right)}\left(1-\frac{0.3}{2}\right)\\
%&=0.32 \times 10^{-3}
\end{aligned}\]
logitudinal strain
\[\begin{aligned}
\varepsilon_{l} &=\frac{p d}{2 t E}\left(\frac{1}{2}-\frac{1}{m}\right)\\
%&=\frac{3 \times 500}{2 \times 10 \times\left(200 \times 10^{3}\right)}\left(\frac{1}{2}-0.3\right)\\
%&=0.075 \times 10^{-3}
\end{aligned}\]
original volume of the vessel
\[\begin{aligned}
V &=\frac{\pi}{4}(d)^{2} \times l \\
%&=\frac{\pi}{4}(500)^{2} \times\left(2 \times 10^{3}\right)\\
%&=392.7 \times 10^{6} \mathrm{~mm}^{3}
\end{aligned}\]
\(\therefore \quad\) Change in volume,
\[\begin{aligned}
\delta V &=V\left(\varepsilon_{c}+2 \varepsilon_{l}\right)\\
%&=392.7 \times 10^{6}\left[0.32 \times 10^{-3}+\left(2 \times 0.075 \times 10^{-3}\right)\right] \mathrm{mm}^{3} \\
%&=\mathbf{1 8 5} \times \mathbf{1 0}^{3} \mathbf{m m}^{3}
\end{aligned}\]
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Example 2 |
A cylindrical thin drum \(800\ mm\) in diameter and \(4\ m\) long is made of \(10\ mm\) thick plates. The drum is subjected to an internal pressure of \(2.5\ MPa\). Taking \(E\) as \(200\ GPa\) and Poisson’s ratio as \(0.25\), calculate the change in diameter and length. |
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Solution:
change in diameter
\[\begin{aligned}
\delta d &=\frac{p d^{2}}{2 t E}\left(1-\frac{1}{2 m}\right) \
%&=\frac{2.5 \times(800)^{2}}{2 \times 10 \times\left(200 \times 10^{3}\right)}\left(1-\frac{0.25}{2}\right) \mathrm{mm} \
%&=\mathbf{0 . 3 5} \mathbf{~ m m}
\end{aligned}\]
change in length \begin{aligned} \delta l &=\frac{p d l}{2 t E}\left(\frac{1}{2}-\frac{1}{m}\right)\ %&=\frac{2.5 \times 800 \times\left(4 \times 10^{3}\right)}{2 \times 10 \times\left(200 \times 10^{3}\right)}\left(\frac{1}{2}-0.25\right) \mathrm{mm} \ %&=\mathbf{0 . 5} \mathbf{~ m m} \end{aligned} |
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Example 3 |
A cylindrical shell of diameter \(500\ mm\) is required to withstand an internal pressure of \(4\ MPa\). Find the minimum thickness of the shell, if the maximum tensile strength of the plate material is \(400\ MPa\) and the efficiency of the joints is \(65\%\). Take the factor of safety as \(5\). |
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Solution:
allowable tensile stress (i.e., circumferential stress),
\[\begin{aligned}
\sigma_{c} &=\frac{\text { Tensile strength }}{\text { Factor of safety }}\
%&=\frac{400}{5}=80 \mathrm{~N} / \mathrm{mm}^{2} \
\end{aligned}\]
and minimum thickness of shell
\[\begin{aligned}
t&=\frac{p d}{2 \sigma_{c} \eta}\
%&=\frac{4 \times 500}{2 \times 80 \times 0.65}=\mathbf{1 9 . 2} \mathbf{~ m m}\
%&\approx 20 \mathrm{~mm}
\end{aligned}\]
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Example 4 |
A cylindrical shell of \(1.3\ m\) diameter is made up of \(18\ mm\) thick plates. It is subjected to an internal pressure of \(2.4\ MPa\). Taking the efficiency of the joints to be \(70\%\), calculate the circumferential and longitudinal stress. |
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Solution:
circumferential stress \begin{aligned} \sigma_{c} &=\frac{p d}{2 t \eta} \ %&=\frac{2.4 \times\left(1.3 \times 10^{3}\right)}{2 \times 18 \times 0.7}=124 \mathrm{~N} / \mathrm{mm}^{2}\ %&=\mathbf{1 2 4} \text { MPa } \end{aligned} longitudinal stress, \begin{aligned} \sigma_{l} &=\frac{p d}{4 t \eta}\ %&=\frac{2.4 \times\left(1.3 \times 10^{3}\right)}{4 \times 18 \times 0.7}=62 \mathrm{~N} / \mathrm{mm}^{2}\ %&=\mathbf{6 2} \mathbf{M P a} \end{aligned} |
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Example 5 |
A cylindrical shell, \(0.8 \mathrm{~m}\) in diameter and \(3 \mathrm{~m}\) long is having \(10 \mathrm{~mm}\) wall thickness. If the shell is subjected to an internal pressure of \(2.5 \mathrm{~N} / \mathrm{mm}^{2}\), taking \(E=200 \mathrm{GPa}\) and Poisson’s ratio \(=0.25\), determine: change in diameter, change in length, and change in volume. |
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Solution:
\[\begin{aligned}
\text{Hoop stress, } \sigma_{n}&=\frac{p d}{2 t}\\
\text{Longitudinal stress, } \sigma_{l}&=\frac{p d}{4 t}\\
\text{Hoop strain, } \varepsilon_{n}&=\frac{1}{E}\left(\sigma_{n}-\varepsilon \sigma_{l}\right) \\
\text{Longitudinal strain, } \varepsilon_{l}&=\frac{1}{E}\left(\sigma_{l}-\varepsilon \sigma_{n}\right) \\
\text{Volumetric strain, } \varepsilon_{v}&=2 \varepsilon_{n}+\varepsilon_{l} \\
\end{aligned}
\begin{aligned}
\text{Increase in diameter } &= \text{Hoop strain } \times \text{original diameter}\\
\text{Increase in length } &= \text{Longitudinal strain } \times \text{original length}\\
\text{Increase in volume } &= \text{Volumetric strain } \times \text{original volume}\\
\text{Original volume } &= \frac{\pi d^{2}}{4} \times l \\
\end{aligned}\]
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Example 6 |
A boiler shell is to be mad of \(15 \mathrm{~mm}\) thick plate having a limiting tensile stress of \(120 \mathrm{~N} / \mathrm{mm}^{2}\). If the efficiencies of the longitudinal and circumferential joints are \(70 \%\) and \(30 \%\) respectively determine: the maximum permissible diameter of the shell for an internal pressure of \(2 \mathrm{~N} / \mathrm{mm}^{2}\),and permissible intensity of internal pressure when the shell diameter is \(1.5 \mathrm{~m}\). |
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Solution:
\[\begin{aligned}
\text{Circumferential stress } \sigma_{c} & =\frac{p d}{2 t \times \eta_{1}} \\
\text{Longitudinal stress } \sigma_{l} & =\frac{\mathrm{pd}}{4 \mathrm{t} \times \eta_{\mathrm{c}}} \\
\end{aligned}\]
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Example 7 |
A thin cylindrical shell \(100 \mathrm{~cm}\) diameter, \(1 \mathrm{~cm}\) thick and \(5 \mathrm{~m}\) long when subjected to internal pressure of \(3 \mathrm{~N} / \mathrm{mm} 2\). Take the value of \(E=2 \times 10^5 \mathrm{~N} / \mathrm{mm}^{2}\) and poisons ratio \(\mu=0.3\) Calculate: the change in diameter, change in length, and change in volume of a thin cylindrical shell. |
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Solution:
\[\begin{aligned}
\text{Change in diameter, } \delta \mathrm{d} & =\frac{p d^{2}}{2 t E}\left(1-\frac{\mu}{2}\right) \\
\text{Change in length, } \delta \mathrm{L} & =\frac{p d L}{2 t E}\left(\frac{1}{2}-\mu\right) \\
\text{Change in volume, } \delta \mathrm{V} & =\mathrm{V}\left[2 \mathrm{e}_{1}+\mathrm{e}_{2}\right] \\
& =\mathrm{V}\left[2 \times \frac{\delta \mathrm{d}}{\mathrm{d}}+\frac{\delta L}{L}\right] \\
\mathrm{V}= \text{original volume } &=\frac{\pi}{4} \times d^{2} \times \mathrm{L}
\end{aligned}\]
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Example 8 |
A round shell \(90 \mathrm{~cm}\) long \(20 \mathrm{~cm}\) internal diameter having thickness of metal as \(8 \mathrm{~mm}\) is filled with fluid at atmospheric pressure. If an additional \(20 \mathrm{~cm}^{3}\) of fluid is pumped into the cylinder, and taking \(\mathrm{E}=2 \times 10^{5} \mathrm{~N} / \mathrm{mm}^{2}\) and \(\mu=0.3\), find: the volume of the cylinder, the pressure exerted by the fluid on the cylinder, and the hoop stress induced. |
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Solution:
\[\begin{aligned}
\delta \mathrm{V} &=\text { Volume of additional fluid }\\
\text{Volume of cylinder } \mathrm{V} &=\frac{\pi}{4} \times d^{2} \times \mathrm{L} \\
\text{Let } \mathrm{p} &= \text{ pressure exerted by fluid on the cylinder} \\
\text{Now using volumetric strain, } \frac{\delta V}{V} &=2 \mathrm{e}_{1}+\mathrm{e}_{2} \\
\end{aligned}\\
\text{But } \mathbf{e}_{1} \text{ and } \mathbf{e}_{2} \text{ are circumferential and longitudinal strains, respectively.}\\
\begin{aligned}
\mathrm{e}_{1} & =\frac{p d}{2 t E}\left(1-\frac{\mu}{2}\right) \\
\mathrm{e}_{2} & =\frac{p d}{2 t E}\left(\frac{1}{2}-\mu\right) \\
\text{Hoop stress } \sigma_{1} &=\frac{p d}{2 t}
\end{aligned}\]
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5.2. A Case Study of the OceanGate Titan Incident
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Assignment 1: From the Titanic to the OceanGate Titan: Implosion Analysis and Design Improvement of Thin-Walled Cylinders |
Introduction The OceanGate Titan incident serves as a devastating reminder of the significance of thin-walled pressure vessel design in Solid Mechanics. For this assignment, your task is to explore the determinants affecting the implosion pressure of a thin-walled cylinder composed of carbon fiber, while examining potential design modifications that might have circumvented the OceanGate Titan disaster. Instructions - Identify the factors impacting the implosion pressure of a thin-walled cylinder manufactured from carbon fiber. - Examine the plausible causes of the OceanGate Titan incident, considering these factors. - Propose design modifications, based on the principles of thin-walled pressure vessel design, that might have prevented the OceanGate Titan disaster. - Compile your findings in one A4 page. Grading Your assignment will be assessed based on the following criteria: - Comprehensiveness of your analysis pertaining to solid mechanics. - Validity of your proposed design changes. - Clarity and accuracy of your written communication, aligning with scientific facts. |
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Solution:
Please attempt this assignment. Additional Information On June 23, 2023, a deep-sea submersible named the OceanGate Titan vanished during a commercial expedition to the Titanic wreckage. The vessel was carrying five individuals, including OceanGate’s CEO, Stockton Rush, none of whom survived the incident. The exact cause of the disaster remains under investigation, with multiple explanations proposed. One theory suggests a hull breach leading to the flooding and implosion of the submersible. Another theory posits a malfunction within the submersible’s battery system leading to a fire or explosion. The OceanGate Titan was constructed from carbon fiber, an anisotropic material with properties that vary depending on the direction of the applied load. This complicates testing procedures, as conventional methods are based on isotropic materials' assumptions. Furthermore, potential points of weakness could exist at the joints where different materials meet, possibly contributing to the implosion. The OceanGate Titan disaster eerily mirrors the infamous 1912 Titanic accident. While the latter hit an iceberg due to certain navigational and design flaws, the former suffered an implosion, likely due to structural or material failures. This incident underscores the inherent risks associated with deep-sea diving. Regardless of experience level, all divers must recognize the potential dangers and conduct thorough risk assessment and preparation prior to embarking on commercial deep-sea diving expeditions. |
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Comments (7)
Ae we doing chapters 6, 7, 8 & 9?
No. That's for Solid II.
Kindly confirm for us the formula for volumetric strain
Hello, are the notes here all we will have for SSM2 or is there a possibility for addition?
Solid Mechanics II starts from chapter 6.
Could you confirm if in example 1 the shear force diagram is correct since from my shear force calculations the forces are positive meaning the diagram ought to be upwards
Yes, in example 1, the shear force diagram is correct. There are two point loads acting downwards, which tend to shear the bar in the negative direction.